What quantity of aluminum is deposited when a current of 10A is passed through a solution of an aluminum salt for 1930s?
[Al = 27, F = 96500 C mol -1]
0.2 g
1.8 g
5.4 g
14.2 g
Explanation
Video Explanation
Post your Contribution
Discussions (21)
Q=It
Q=10x1930=19300Coulombs.
1 Faraday=96500Coul.
Al³+
27g
3 X 96500C liberate 27g of Al
19300 liberates Xg
cross multiply
19300x27=3 X 96500 x X
521100=289500X
521100
X= ————
289500
X=1.8g
.
. .
quantity of electricity=1930×10=19300
27g of Al³+ will be liberated by 3×96500=289500
27×19300÷289500=1.8g.....is dat clear
also using a very straight forward formula
M= Rmm*I*t/ n*F
where Rmm= 27
I= 10A
t= 1930s
n= +3(bcoz aluminum has a +3 charge)
F= 96500
M= 27*10*1930/ 3*96500
M= 521100/289500
M= 1.8g
Hope this helps
M=ZIT
l=10A
T=1930
Z=27÷(3×96500)
=27÷289500=0.000093
M=0.000093×10×1930
M=1.79
M=1.8
Q=It
Q=10X19300
Q=19300C
Al^(3+) + 3F:Al
27g. 3(96500):27g
Aluminum deposited is 27X19300/3(96500)
=1.8g
Q=it
Q=10×1930
Q=19,300
Al^3+ +3e--------> Al
1 mole ------> 3F
x mole-------> 19,300
x=0.07
mole=mass/molar mass
0.07=mass/27
mass=0.07×27
mass=1.8g
OR
mass=molar mass × Quantity/Charge × Faraday
m= Mm×Q/e×f
m= 27×19300/3×96500
m=1.8g
Q=It...10*1930=19300c...1m=96500c,19300c=0.2m...Al=+3...3m=27g,0.2m=???0.2*27|3=1.8g...To my understanding though
okay for those who don't understand lemme explain
Quantity of electricity(Q) = Current(I) x Time(t)
= 10 × 1930
3F will displace 27g of aluminium(27g is the mass of aluminium) F=96500
3 × 96500 will displace 27g of aluminium
10 ×1930 will displace x gram of aluminium
so cross multiply
289500x =19300 × 27
x = 19300 × 27÷(289500)
x = 1.8g
